A central limit theorem for fluctuations in 1D stochastic homogenization

Abstract

In this paper, we analyze the random fluctuations in a 1D stochastic homogenization problem and prove a central limit theorem: the first order fluctuations is described by a Gaussian process that solves an SPDE with an additive spatial white noise. Using a probabilistic approach, we obtain a precise error decomposition up to the first order, which also helps to decompose the limiting Gaussian process, with one of the components corresponding to the corrector obtained by a formal two-scale expansion.

Appendix: Technical lemmas

Lemma 6.1

$\mathbb {E}\{|\tilde{\phi }(x)|^2\}\lesssim |x|$ and $\mathbb {E}\{|\tilde{\psi }(x)|^2\}\lesssim |x|^3$.

Proof

Since $\tilde{\phi }(x)=\bar{a}\int _0^x \tilde{V}(y)dy$ and R(x) is the integrable covariance function of $\tilde{V}$, we have

$$\begin{aligned} \mathbb {E}\big \{|\tilde{\phi }(x)|^2\big \}\lesssim \int _0^x\int _0^x R(y-z)dydz \lesssim |x|. \end{aligned}$$

For $\tilde{\psi }(x)$, by (3.11) we have

$$\begin{aligned} \tilde{\psi }(x)=-\frac{2}{\bar{a}}\int _0^x \tilde{\phi }(y)\big (\tilde{V}(y)+\bar{a}^{-1}\big )dy=-2\int _0^x(\tilde{V}(y)+\bar{a}^{-1})\int _0^y \tilde{V}(z)dzdy, \end{aligned}$$

so

$$\begin{aligned} \begin{aligned}&\mathbb {E}\{|\tilde{\psi }(x)|^2\}\\&\quad \lesssim \int _0^x\int _0^x\int _0^{y_1}\int _0^{y_2}|\mathbb {E}\{(\tilde{V}(y_1)+\bar{a}^{-1})(\tilde{V}(y_2)+\bar{a}^{-1})\tilde{V}(z_1)\tilde{V}(z_2)\}|dz_1dz_2dy_1dy_2. \end{aligned} \end{aligned}$$

In the above expression, we need to control the second, third and fourth moments of $\tilde{V}$, which is a mean-zero stationary random field of finite range dependence. For the term with the second moment, we have

$$\begin{aligned} \int _0^x\int _0^x \int _0^{y_1}\int _0^{y_2}|R(z_1-z_2)|dz_1dz_2dy_1dy_2\lesssim |x|^3. \end{aligned}$$

The other cases are discussed in the same way by applying Lemma 6.2. $\square$

Lemma 6.2

(Moment estimates) For any $x_i\in \mathbb {R}, i=1,2,3,4$, we have

$$\begin{aligned} |\mathbb {E}\left\{ \prod _{i=1}^3 \tilde{V}(x_i)\right\} |\le \rho (|x_1-x_2|)+\rho (|x_1-x_3|)+\rho (|x_2-x_3|), \end{aligned}$$

(5.9)

and

$$\begin{aligned} \begin{aligned} |\mathbb {E}\left\{ \prod _{i=1}^4 \tilde{V}(x_i)\right\} |&\le \rho (|x_1-x_2|)\rho (|x_3-x_4|)+\rho (|x_1-x_3|)\rho (|x_2-x_4|)\\&\quad +\rho (|x_1-x_4|)\rho (|x_2-x_3|) \end{aligned} \end{aligned}$$

(5.10)

for some $\rho :\mathbb {R}_+\rightarrow \mathbb {R}_+$ satisfying $\rho (r)\lesssim 1\wedge r^{-p}$ for any $p>0$.

Proof

Since $\tilde{V}$ is bounded, mean zero and of finite range dependence, (5.10) comes from [1, Lemma 3.1]. For (5.9), it is clear that there exists a compactly supported $\rho :\mathbb {R}_+\rightarrow \mathbb {R}_+$ such that

$$\begin{aligned} \begin{aligned} |\mathbb {E}\left\{ \prod _{i=1}^3 \tilde{V}(x_i)\right\} |&\le \rho (\min \{ |x_1-x_2|,|x_1-x_3|,|x_2-x_3|\})\\ \le&\rho (|x_1-x_2|)+\rho (|x_1-x_3|)+\rho (|x_2-x_3|). \end{aligned} \end{aligned}$$

The proof is complete. $\square$

Lemma 6.3

(Estimates on local time) Let $L_t^x(y)$ be the local time of a standard Brownian motion $W_t$ starting from x up to t, then for any $p\ge 1$,

$$\begin{aligned} \mathbb {E}\{|L_t^x(y)|^p\}\lesssim t^{\frac{p}{2}}\int _{|y-x|}^\infty q_t(z)dz. \end{aligned}$$

Proof

First, $L_t^x(y)$ has the same distribution as $L_t^0(y-x)$. By the strong Markov property of Brownian motion and distribution property of $L_t^0(0)$, we further have

$$\begin{aligned} L_t^0(y-x)\sim L_{t-\tau _{y-x}}^0(0)1_{\tau _{y-x}\le t}\sim M_{t-\tau _{y-x}}1_{\tau _{y-x}\le t}, \end{aligned}$$

where $\tau _{y-x}$ is the hitting time of another independent Brownian motion starting at zero and reaching at $y-x$, and $M_t$ is the maximum of $W_t$ during [0, t]. Thus we have

$$\begin{aligned} \begin{aligned} \mathbb {E}\{|L_t^x(y)|^p\}=\int _0^t \mathbb {E}\{ |M_{t-s}|^p\} p^{\tau _{y-x}}(s)ds\lesssim&t^{\frac{p}{2}}\int _0^t p^{\tau _{y-x}}(s)ds\\ =&\,\, t^{\frac{p}{2}} {\mathbb {P}}(\tau _{y-x}\le t), \end{aligned} \end{aligned}$$

with $p^{\tau _{y-x}}$ the density of $\tau _{y-x}$. The reflection principle tells that

$$\begin{aligned} {\mathbb {P}}(\tau _{y-x}\le t)=2\int _{|y-x|}^\infty q_t(z)dz. \end{aligned}$$

The proof is complete. $\square$

Lemma 6.4

(SPDE representation) Let $v(t,x)=\mathbb {E}_W\{ f(x+W_t)\int _0^t \dot{\mathcal {W}}(x+W_s)ds\}$, then it solves

$$\begin{aligned} \partial _t v(t,x)=\frac{1}{2}\partial _{x}^2v(t,x)+ u(t,x)\dot{\mathcal {W}}(x) \end{aligned}$$

(5.11)

with zero initial condition, and the function u solving $\partial _t u=\frac{1}{2}\partial _{x}^2 u$ with initial condition $u(0,x)=f(x)$.

Proof

The proof is similar to that of Proposition 5.1. First, we approximate the SPDE with a smooth equation. Then we use the probabilistic representation of the smooth equation and show its convergence.

The solution to (5.11) can be written as

$$\begin{aligned} v(t,x)= & {} \int _0^t \int _\mathbb {R}q_{t-s}(x-y)u(s,y)\mathcal {W}(dy)ds\\= & {} \int _\mathbb {R}\left( \int _0^tq_{t-s}(x-y)u(s,y)ds\right) \mathcal {W}(dy), \end{aligned}$$

and we define $v_\varepsilon (t,x)$ as

$$\begin{aligned} v_\varepsilon (t,x)= & {} \int _0^t \int _\mathbb {R}q_{t-s}(x-y)u(s,y)\mathcal {W}_\varepsilon (y)dyds\\= & {} \int _\mathbb {R}\left( \int _0^tq_{t-s}(x-y)u(s,y)ds\right) \mathcal {W}_\varepsilon (y)dy, \end{aligned}$$

with

$$\begin{aligned} \mathcal {W}_\varepsilon (y)=\int _\mathbb {R}\frac{1}{\varepsilon }h\Big (\frac{x-y}{\varepsilon }\Big )\mathcal {W}(dy) \end{aligned}$$

as a smooth mollification of $\mathcal {W}$. It is clear that $v_\varepsilon (t,x)\rightarrow v(t,x)$ in $L^2(\Omega )$ as $\varepsilon \rightarrow 0$.

Since $v_\varepsilon$ solves the equation

$$\begin{aligned} \partial _t v_\varepsilon =\frac{1}{2}\partial _x^2 v_\varepsilon +u \mathcal {W}_\varepsilon , \end{aligned}$$

by a probabilistic representation we can rewrite the solution as

$$\begin{aligned} v_\varepsilon (t,x)=\mathbb {E}_W\left\{ \int _0^t u(t-s,x+W_s)\mathcal {W}_\varepsilon (x+W_s)ds\right\} . \end{aligned}$$

Since u solves the heat equation with initial condition $u(0,x)=f(x)$, we obtain

$$\begin{aligned} \begin{aligned} v_\varepsilon (t,x)&=\mathbb {E}_W\mathbb {E}_B\left\{ \int _0^t f(x+W_s+B_{t-s})\mathcal {W}_\varepsilon (x+W_s)ds\right\} \\&= \mathbb {E}_W\left\{ f(x+W_t)\int _0^t \mathcal {W}_\varepsilon (x+W_s)ds\right\} \\&= \mathbb {E}_W\left\{ f(x+W_t)\int _\mathbb {R}\mathcal {W}_\varepsilon (y) L_t^x(y)dy\right\} , \end{aligned} \end{aligned}$$

where $L_t^x(y)$ is the local time of $x+W_t$.

By Lemma 6.3, for any $p\ge 1$, $\mathbb {E}\{|L_t^x(y)|^p\}$ can be bounded by some integrable function in y, and this helps to pass to the limit

$$\begin{aligned}&v_\varepsilon (t,x)\rightarrow \mathbb {E}_W\left\{ f(x+W_t)\int _\mathbb {R}L_t^x(y)\mathcal {W}(dy)\right\} \\&\quad =\mathbb {E}_W\left\{ f(x+W_t)\int _0^t \dot{\mathcal {W}}(x+W_s)ds\right\} \end{aligned}$$

in $L^2(\Omega )$. The proof is complete. $\square$